If I_{n+1} = int_{0}^{1} frac{x^{n+1} - 1}{x | vedclass

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(D) Given $I_{n+1} = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx$.Consider $I_{n+1} + I_n = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx + \int_{0}^{1} \fr

Vedclass · Accepted Answer

$\frac{1}{11}$

Vedclass · Answer

(D) Given $I_{n+1} = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx$.Consider $I_{n+1} + I_n = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx + \int_{0}^{1} \frac{x^n - 1}{x + 1} dx$.$I_{n+1} + I_n = \int_{0}^{1} \frac{x^{n+1} + x^n - 2}{x + 1} dx$.$I_{n+1} + I_n = \int_{0}^{1} \frac{x^n(x + 1) - 2}{x + 1} dx$.$I_{n+1} + I_n = \int_{0}^{1} x^n dx - 2 \int_{0}^{1} \frac{1}{x + 1} dx$.$I_{n+1} + I_n = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 - 2 [\log(x + 1)]_0^1$.$I_{n+1} + I_n = \frac{1}{n+1} - 2 \log 2$.For $n = 10$,we have $I_{11} + I_{10} = \frac{1}{11} - 2 \log 2$.Therefore,$I_{10} + I_{11} + 2 \log 2 = \frac{1}{11}$.

If $I_{n+1} = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx$,then the value of $I_{10} + I_{11} + 2 \log 2$ is:

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